PE 439 【莫比乌斯反演】【杜教筛】

垃圾题，坑了我一天QAQ

题目大意：求 ∑Ni=1∑Nj=1σ(ij) , N=1011

∑Ni=1∑Nj=1σ(ij)
= ∑Ni=1∑Nj=1∑x|i∑y|jxjy[gcd(x,y==1)]
= ∑Ni=1∑Nj=1∑x|i∑y|jxjy∑d|x,yμ(d)
= ∑Nd=1μ(d)∑⌊nd⌋i=1id⌊nid⌋∑⌊nd⌋j=1j⌊nid⌋
= ∑Nd=1dμ(d)(∑⌊nd⌋i=1σ(i))2

这貌似跟SDOI2015的那道题差不多，但是这里有个 jy ,就半天没有想出来QAQ
对于 ∑Ni=1σ(i) ，前 N23 项可以线筛预处理，后面可以暴力分块求

然后是处理 ∑Ni=1iμ(i) ,然后又卡了TAT。。。
这就是一个杜教筛啊QAQ

g(1)S(n)=∑ni=1(f∗g)(i)−∑i=2g(i)S(⌊ni⌋)
这里 f(n)=iμ(i) , g(n)=n
就有 S(n)=1−∑ni=2iS(⌊ni⌋)
前 N23 项预处理，后面记忆化搜索

复杂度 O(N23)

【答案】968697378

#include<iostream>
#include<algorithm>
#include<ctime>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cstdio>
#define N 21543983
#define M 1000000000
#define LL long long
using namespace std;

LL n = (LL)1e11,tot,siz;
LL s0[N],pr[N],num[N],f[N],miu[N];
int prime[N];
bool v[N];
LL first[N],to[N],next[N],S[N];

void get_prime()
{
    num[1] = 1;
    miu[1] = 1;
    f[1] = 1;
    for (int i = 2;i < N;i ++)
    {
        if (!v[i])
        {
            prime[tot ++] = i;
            miu[i] = -1;
            num[i] = 1 + i;
            s0[i] = 1 + i;
            pr[i] = 1;
        }
        for (int j = 0;j < tot && i * prime[j] < N;j ++)
        {
            v[i * prime[j]] = true;
            if (i % prime[j] == 0)
            {
                pr[i * prime[j]] = pr[i];
                s0[i * prime[j]] = (s0[i] * prime[j] + 1) % M;
                num[i * prime[j]] = pr[i] * s0[i * prime[j]] % M;
                break;
            }
            else 
            {
                miu[i * prime[j]] = -miu[i];
                pr[i * prime[j]] = num[i];
                s0[i * prime[j]] = prime[j] + 1;
                num[i * prime[j]] = num[i] * s0[i * prime[j]] % M;
            }
        }
        f[i] = (f[i - 1] + i * miu[i]) % M;
    }
    for (int i = 2;i < N;i ++)
        (num[i] += num[i - 1]) %= M;
}

LL find(LL n)
{
    LL x = n % N;
    for (LL i = first[x];i;i = next[i])
        if (to[i] == n) return S[i];
    return -1;
}

void inser(LL n,LL w)
{
    if (w < 0) w += M;
    LL x = n % N;
    next[++ siz] = first[x];
    first[x] = siz;
    to[siz] = n;
    S[siz] = w;
}

LL mul2(LL a,LL b)
{
    (~a & 1) ? a >>= 1 : b >>= 1;
    return ((a % M) * (b % M)) % M;
}

LL cal(LL n)
{
    if (n < N) return f[n];
    LL S = find(n);
    if (~ S) return S;
    LL ret = 1;
    for (LL i = 2,j;i <= n;i = j + 1)
    {
        j = min(n,n / (n / i));
        (ret -= mul2(j + i,j - i + 1) * cal(n / i)) %= M;
    }
    inser(n,ret);
    return ret;
}

LL sigma(LL n)
{
    if (n < N) return num[n];
    LL ret = 0;
    for (LL i = 1,j;i <= n;i = j + 1)
    {
        j = min(n,n / (n / i));
        (ret += mul2(j + i,j - i + 1) * (n / i)) %= M;
    }
    return ret;
}

LL pf(LL x)
{
    return x * x % M;
}

int main()
{
    get_prime();
    LL ans = 0;
    for (LL i = 1,j;i <= n;i = j + 1)
    {
        j = min(n,n / (n / i));
        (ans += (cal(j) - cal(i - 1)) * pf(sigma(n / i))) %= M;
    }
    if (ans < 0) ans += M;
    cout << ans << endl;

    return 0;
}